A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
Does the following series converge or diverge?
anonymous
 5 years ago
Does the following series converge or diverge?

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{n=2}^{infinity} 1/n \sqrt{\ln(n)}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm wanting to say it converges to 1, but I'm not sure how to prove that. Real analysis wasn't my strong point.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you know what test to use?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0my original thought was ratio, but i dont know how to apply that here

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm wanting to say it has something to do with the Cauchy Criterion

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0...I have no idea what that means..haha

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I don't think we covered that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, the ratio test should work.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0really? Let me try that really quickly

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, that gave me infinity over infinity...maybe i'm doing it wrong?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Looking up properties of the natural log right now, give me a sec.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Integral test would be easier since i see ln(n) and 1/n there.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok all i can think of is that \[n / (n+1) \le 1\] and \[\sqrt(\ln(n+1)) \le n\] and \[\sqrt(\ln(n)) \le n\] thus \[\left n/(n+1) * \sqrt(\ln(n+1)) / \sqrt(\ln(n)) \right \le \left 1 * n/n \right \le 1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmm. Ok, let me think about that for a min :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmm actually that might not work cause the ration needs to be less than 1 for the ratio test to work

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when i worked it doing the integral test i got infinity meaning divergent does that seem right?
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.